You have found the following ages (in years) of 6 lions. Those lions were randomly selected from the 33 lions at your local zoo: $ 8,\enspace 4,\enspace 4,\enspace 1,\enspace 15,\enspace 2$ Based on your sample, what is the average age of the lions? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 33 lions, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\overline{x}} = \dfrac{8 + 4 + 4 + 1 + 15 + 2}{{6}} = {5.7\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {5.29} + {2.89} + {2.89} + {22.09} + {86.49} + {13.69}} {{6 - 1}} $ {s^2} = \dfrac{{133.34}}{{5}} = {26.67\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{26.67\text{ years}^2}} = {5.2\text{ years}} $ We can estimate that the average lion at the zoo is 5.7 years old. There is also a standard deviation of 5.2 years.